Question

If $y = 1 - x + \dfrac{{{x^2}}}{{2!}} - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} - .....$, then $\dfrac{{{d^2}y}}{{d{x^2}}}$ is equal to
$\eqalign{
  & 1) - x \cr
  & 2)x \cr
  & 3)y \cr
  & 4) - y \cr} $

Answer 1

Hint: In the given question, we need to find $\dfrac{{{d^2}y}}{{d{x^2}}}$, which is double differentiation or differentiating twice. There are factorials of $2$ and $3$ in the question, we do not have to find the values. While differentiating once, we only simplify such terms. Then we differentiate it again.
The formulas used to solve the problem are:
$\eqalign{
  & \dfrac{d}{{dx}}const = 0 \cr
  & \dfrac{d}{{dx}}\left( x \right) = 1 \cr
  & \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{\left( {n - 1} \right)}} \cr} $

Complete step-by-step answer:
The given expression is as follows,
$y = 1 - x + \dfrac{{{x^2}}}{{2!}} - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} - .....$
We do not know the last term, so we write is as it is given
Now, differentiating with respect x, we get
$\dfrac{{dy}}{{dx}} = - 1 + \dfrac{{2x}}{{2!}} - \dfrac{{3{x^2}}}{{3!}} + \dfrac{{4{x^3}}}{{4!}} - .....$
This can be done using the above-mentioned formulas.
Now, simplifying it further,
$\dfrac{{dy}}{{dx}} = - 1 + x - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} - ....$
Which will be equal do
$\dfrac{{dy}}{{dx}} = - y$
Now, we need to differentiate the above given expression to get $\dfrac{{{d^2}y}}{{d{x^2}}}$
$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - dy}}{{dx}}$
$\eqalign{
  & \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - ( - y) \cr
  & \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = y \cr} $
Therefore, the final answer is $y$.
Hence, option (3) is the correct answer.
So, the correct answer is “Option 3”.

Note: The question asks for double differentiation, even though we get a simple answer after differentiating once, do not forget to repeat the process. After the first differentiation, the answer is negative, so carry the sign as is it for the next step. Be careful about the signs and their derivatives. Learn the formula and apply them appropriately.